Question

If A \(= \;\left[ {\begin{array}{*{20}{c}} 1&{ 1}\\ 4&{ 6} \end{array}} \right]\) , find k so that \({A^2} = kA - 2I\), where I is an identity matrix.
1. 7
2. 8
3. 9
4. 6

Answer 1

Correct Answer - Option 1 : 7

Concept:

 If A and B are two matrices such that the no. of columns of A is equal to the no. of rows of B. If A = [aij] is a m × n matrix and B = [bij] be a n × p matrix, then the product AB is the resultant matrix of order m × p and is defined as:

\({\left( {AB} \right)_{ij}} = \;\mathop \sum \limits_{k = 1}^n {a_{ik}} \times {b_{kj}}\forall \;i = 1,\;2, \ldots ,m\;and\;j = 1,\;2,\; \ldots .,\;p\)

Calculation:

Given: A \(= \;\left[ {\begin{array}{*{20}{c}} 1&{ 1}\\ 4&{ 6} \end{array}} \right]\)

Here, we have to find the value of k such that \({A^2} = kA - 2I\)

\({A^2}\; = \;A.A\; = \;\left[ {\begin{array}{*{20}{c}} 1&{ 1}\\ 4&{ 6} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&{ 1}\\ 4&{ 6} \end{array}} \right]\;\\ = \;\left[ {\begin{array}{*{20}{c}} {1\left( 1 \right)\; + \;\left( { 1} \right)\left( 4 \right)}&{1\left( { 1} \right)\; + \;\left( { 1} \right)\left( { 6} \right)}\\ {4\left( 1 \right)\; + \;\left( { 6} \right)\left( 4 \right)}&{4\left( { 1} \right)\; + \;\left( { 6} \right)\left( { 6} \right)} \end{array}} \right]\; = \;\left[ {\begin{array}{*{20}{c}} 5&{ 7}\\ 28&{ 40} \end{array}} \right]\)

\({A^2}\; = \;kA - 2I\)

\(\left[ {\begin{array}{*{20}{c}} 5&{ 7}\\ 28&{ 40} \end{array}} \right]\; = \;k\left[ {\begin{array}{*{20}{c}} 1&{ 1}\\ 4&{ 6} \end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\\= \;\left[ {\begin{array}{*{20}{c}} {k - 2}&{ k}\\ {4k}&{ 6k - 2} \end{array}} \right]\)

As the two matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements:

4k = 28 ⇒ k = 7

∴ The value of k is 7.